Figure 1 sho ws a circuit that is appropriate for either a current- or a voltage-feedback amp. If the feedback resistance, RF, equals 2RIN, where RIN is the input resistance, the closed-loop gain of each channel is –2V/V. At first glance, it is easy to assume that the closed-loop bandwidth equals the gain-bandwidth product divided by each channel’s gain, or |–2V/V|. Don’t make this assumption!

　　If you use a voltage- or current-feedback amp with the circuit in Figure 1, the noise gain is:

　　where N is the

number of input channels. This circuit’s bandwidth, with a voltage-feedback amp, equals the gain-bandwidth product divided by the noise gain. For instance, if you have a voltage-feedback amp with a gain-bandwidth product of 180 MHz and there are three input channels (N=3) at a gain of –2V/V, the circuit’s closed-loop bandwidth is 25.7 MHz. Additional channels reduce the closed-loop bandwidth, even though the input signals continue to see a gain of –2V/V.

　　If you use a current-feedback amp with the circuit in Figure 1, the amplifier’s closed-loop bandwidth depends less on the closed-loop gain and the number of input channels. If you design this circuit with such an amp, you would first pICk the optimum feedback resistor, per the manufacturer’s specification and the circuit’s noise gain. You would then select the appropriate value for RIN. From this point, if you add channels to the circuit, a small variation in the signal bandwidth and gain peaking in circuit may oCCur. If that scenario is a concern, go back and refine your feedback-resistor selection. For both current- and voltage-feedback amps, the noise gain always equals the result of Equation 1, but you CAN reduce the feedback-resistor value with the current-feedback-amp circuit and get an increase in circuit bandwidth.

        英文原文地址： http://www.edn.com/article/CA6491140.html

上一页  [1] [2]